In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are $50$ . It moves $0.5 m m$ on main scale in one rotation. The diameter of the ball in figure (ii) is:

2.25 m m

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2.20 m m

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1.20 m m

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1.25 m m

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Solution

The correct option is C $1.20 m m$
From figure (i), it is clear that screw gauge has a zero error.
Least count of screw gauge $= \frac{R e a d i n g o f o n e d i v i s i o n o f m a i n s c a l e}{T o t a l n u m b e r o f d i v i s i o n s o f c i r c u l a r s c a l e}$ $= \frac{0.5}{50} = 0.01 m m$
Therefore, diameter of the ball $=$ (M.S. Reading $+$ Least count (Circular division) $-$ zero error)
$= [1 + 0.01 [25] - 5] = 1.20 m m$

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In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are 50. It moves 0.5mm on main scale in one rotation. The diameter of the ball in figure (ii) is:

In a screw gauge, the zero of main scale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are $50$ . It moves $0.5 m m$ on main scale in one rotation. The diameter of the ball in figure (ii) is: